Question

(a) A shipment of 120 fasteners that contains 4 defective fasteners was sent to a manufacturing plant. The quality-control manager at the manufacturing plant randomly selects 5 fasters and inspects them. What is the probability that exactly 1 fastener is defective? (b) The Energy Information Administration (EIA) sampled 900 retail gasoline outlets and reported that, the national average gasoline price for regular-grade gasoline to be $4.113 per gallon with a standard deviation of $0.11 per gallon. Construct a 95% Confidence Interval for mean price of regular-grade gasoline price.

Answer 1

Answer:

a)  Probability that exactly 1 fastener is defective, P(X = 1) = 0.144

b) Confidence interval for mean price, $CI = [4.1058, 4.1202]$

Step-by-step explanation:

a) Total number of fasteners = 120

Number of defective fasteners = 4

Probability of selecting a defective fastener, p = 4/120

p = 0.033

Probability of selecting an undefective fastener, q = 1 - p

q = 1 - 0.033

q = 0.967

5 fasteners were randomly selected, n =5

Probability that exactly one fastener is defective:

$P(X =r) = (nCr) p^r q^{n-r}\\P(X =1) = (5C1) 0.033^1 0.967^{5-1}\\P(X =1) = 0.144$

b) Number of gasoline outlets sampled, n = 900

Average gasoline price, $\bar{x} = 4.113$

Standard deviation, $\sigma = 0.11$

Confidence Level, CL = 95% = 0.95

Significance level, $\alpha = 1 - 0.95 = 0.05$

$\alpha/2 = 0.05/2 = 0.025$

From the standard normal table, $z_{\alpha/2} = z_{0.025} = 1.96$

error margin can be calculated as follows:

$e_{margin} = z_{\alpha/2} * \frac{\sigma}{\sqrt{n} } \\e_{margin} = 1.96 * \frac{0.11}{\sqrt{900} }\\e_{margin} = 0.0072$

The confidence interval will be given as:

$CI = \bar{x} \pm e_{margin} \\CI = 4.113 \pm 0.0072\\CI = [(4.113-0.0072), (4.113+0.0072)]\\CI = [4.1058, 4.1202]$